second order taylor polynomial

• Nov 4th 2007, 10:03 AM
dopi
second order taylor polynomial
Hi i was wondering if anyone could help me with this question.

i need to use the second order taylor polynomial with remainder to estimate sqraure-root of 101 and give an estimate of the error mode.

but where i get confused is that i am using the formula for the taylor poly..and i dont know what f(a) so i cant dfferentiate and so i cant fill in the taylor poly formula..all i am is given is square root of 101..

• Nov 4th 2007, 10:41 AM
ThePerfectHacker
Quote:

Originally Posted by dopi
Hi i was wondering if anyone could help me with this question.

i need to use the second order taylor polynomial with remainder to estimate sqraure-root of 101 and give an estimate of the error mode.

but where i get confused is that i am using the formula for the taylor poly..and i dont know what f(a) so i cant dfferentiate and so i cant fill in the taylor poly formula..all i am is given is square root of 101..

Let $\displaystyle f(x) = \sqrt{x+100} = (x+100)^{1/2}$ the Taylor polynomial centered at zero or order 2 is, $\displaystyle T_2(x) = 10 + \frac{x}{20} - \frac{x^2}{4000}$ consider the interval $\displaystyle (-2,2)$ the derivative is $\displaystyle f'(x) = \frac{1}{2}(x+100)^{-1/2}$ which is bounded by $\displaystyle \frac{1}{2}(98)^{1/2}$.
So the maximum error possible,
$\displaystyle \frac{\sqrt{98}}{2\cdot 3!}$
• Mar 26th 2009, 08:11 AM
varkoume.com
Hello I have the same problem and I cannot see how you estimated the square root (101)
• Mar 27th 2009, 03:52 AM
mr fantastic
Quote:

Originally Posted by varkoume.com
Hello I have the same problem and I cannot see how you estimated the square root (101)

The exact value is $\displaystyle f(1)$ and the estimate is $\displaystyle T_2 (1)$.